# Why is the heaviside function in the inverse Laplace transform of 1?

• Blanchdog
In summary, the Laplace transform of a function f(t), defined on the real numbers, is ##\int_0^\infty f(t)e^{-st}dt##. The Heaviside function, H(x), is defined to be 0 for x < 0, 1/2 for x = 0, and 1 for x > 0. For the interval in the Laplace integral, the Heaviside function is identical to the constant function f(x) = 1. However, in the context of a larger problem involving the inverse Laplace transform of 1/s, the Heaviside function is multiplied by the inverse transform of 1/s with t substituted by (t-c), in thisf

#### Blanchdog

Homework Statement:: Why is the heaviside function in the inverse laplace transform of 1?
Relevant Equations:: N/A

This is a small segment of a larger problem I've been working on, and in my book it gives the transform of 1 as 1/s and vice versa. But as I've looked online for help in figuring parts of this out, I keep seeing that the heaviside function is the inverse laplace transform of 1/s, when I would think (according to my book) that it should just be 1. Can anyone explain this?

Homework Statement:: Why is the heaviside function in the inverse laplace transform of 1?
Relevant Equations:: N/A

This is a small segment of a larger problem I've been working on, and in my book it gives the transform of 1 as 1/s and vice versa. But as I've looked online for help in figuring parts of this out, I keep seeing that the heaviside function is the inverse laplace transform of 1/s, when I would think (according to my book) that it should just be 1. Can anyone explain this?
The Laplace transform of a function f(t), defined on the real numbers, is ##\int_0^\infty f(t)e^{-st}dt##. The Heaviside function, H(x), is defined to be 0 for x < 0, 1/2 for x = 0, and 1 for x > 0. For the interval in the Laplace integral, the Heaviside function is identical to the constant function f(x) = 1.

The Laplace transform of a function f(t), defined on the real numbers, is ##\int_0^\infty f(t)e^{-st}dt##. The Heaviside function, H(x), is defined to be 0 for x < 0, 1/2 for x = 0, and 1 for x > 0. For the interval in the Laplace integral, the Heaviside function is identical to the constant function f(x) = 1.

Okay, so in the context of the broader problem: $$x=L^{-1} \frac{e^(-2s)}{s(s+2)^2}$$ In which I get the result $$x = \frac{H(t-2)}{4}(1 -e^{4-2t}+2te^{4-2t}-4e^{4-2t})$$ The first term could be replaced by ##H(t-2)## and both would be correct?

Okay, so in the context of the broader problem: $$x=L^{-1} \frac{e^(-2s)}{s(s+2)^2}$$ In which I get the result $$x = \frac{H(t-2)}{4}(1 -e^{4-2t}+2te^{4-2t}-4e^{4-2t})$$ The first term could be replaced by ##H(t-2)## and both would be correct?
I'm not sure what you're doing above.
$$\mathcal L^{-1}[\frac{e^{-2s}}{s(s + 2)^2}] = \mathcal L^{-1}[ \frac{1/4e^{-2s}}{s} + \frac{-1/4e^{-2s}}{s+2} + \frac{-1/2e^{-2s}}{(s + 2)^2}]$$
I don't know why you have four terms inside the parentheses or why you have H(t-2) multiplying those four terms.

I'm not sure what you're doing above.
$$\mathcal L^{-1}[\frac{e^{-2s}}{s(s + 2)^2}] = \mathcal L^{-1}[ \frac{1/4e^{-2s}}{s} + \frac{-1/4e^{-2s}}{s+2} + \frac{-1/2e^{-2s}}{(s + 2)^2}]$$
I don't know why you have four terms inside the parentheses or why you have H(t-2) multiplying those four terms.
I broke up the last term because it was multiplied by ##(t-1)##. I also got the last term as being positive rather than negative in my partial fraction expansion.

EDIT: Found my algebra error, your partial fraction expansion is right.

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I didn't work out the inverse Laplace transforms of what I showed. All I did was to get the partial fraction decomposition.
$$\frac 1 {s(s + 2)^2} = \frac {1/4}{s} + \frac{-1/4}{s+ 2} + \frac{-1/2}{(s + 2)^2}$$
I've checked my work, so I'm pretty confident of the coefficients.

I don't understand what you're saying about being multiplied by t - 1. I also don't understand why you have H(t - 2) multiplying all of your terms.

I've used the theorem ##L (H(t-c)*f(t-c)) = e^{-cs} F(s)##, where the partial fraction expansion above is ##F(s)##, and ##c = 2##. Since the problem is of the form of this theorem, the inverse transform results in a heaviside function multiplied by the inverse transform of ##F(s)## with t substituted by ##(t-c)##, in this case ##(t-2)## (not 1, I misread my work before).

I've used the theorem L(H(t−c)∗f(t−c))=e−csF(s), where the partial fraction expansion above is F(s), and c=2. Since the problem is of the form of this theorem, the inverse transform results in a heaviside function multiplied by the inverse transform of F(s) with t substituted by (t−c), in this case (t−2) (not 1, I misread my work before).
Be careful when you write ##\mathcal L[H(t - c) * f(t - c)]##. I took this at first to mean the transform of the convolution of H(t - c) and f(t - c). Not that the symbol '*' is used for convolution.
Some definitions:
$$f(t) * g(t) = \int_0^1 f(t - x)g(x)dx$$
$$\mathcal L[f(t) * g(t)] = F(s)G(s)$$

Regarding your problem, with the wrinkles removed, do you still have any questions?

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When you take a Laplace Transform of a function ##f##, you typically ignore what ##f(t)## is for negative values of ##t##, so when the table says ##f(t)=1##, what it really means is ##f(t)=1## for ##t\ge 0##. But when you shift this function to the right, the value of ##f(t)## for negative values of ##t## becomes important, and it's assumed that ##f(t)=0## for ##t<0##. So the full definition of ##f## is really
$$f(t) = \begin{cases} 1 & t \ge 0 \\ 0 & t<0 \end{cases},$$ which you should recognize is one of the possible definitions for the Heaviside step function.

On a related note, when you invert, say, ##F(s) = \frac 1{s+a}## and get ##f(t) = e^{-at}##, it's understood that this result is valid only for ##t\ge 0##. Technically, you should write ##f(t) = e^{-at} H(t)##, but pedantically including the step function can get tedious, so it's often omitted and assumed that the reader knows it's there.